Tasks Details
easy
1.
Nesting
Determine whether a given string of parentheses (single type) is properly nested.
Task Score
100%
Correctness
100%
Performance
100%
A string S consisting of N characters is called properly nested if:
- S is empty;
- S has the form "(U)" where U is a properly nested string;
- S has the form "VW" where V and W are properly nested strings.
For example, string "(()(())())" is properly nested but string "())" isn't.
Write a function:
int solution(string &S);
that, given a string S consisting of N characters, returns 1 if string S is properly nested and 0 otherwise.
For example, given S = "(()(())())", the function should return 1 and given S = "())", the function should return 0, as explained above.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [0..1,000,000];
- string S is made only of the characters '(' and/or ')'.
Copyright 2009–2024 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used C++
Time spent on task 3 minutes
Notes
not defined yet
Task timeline
Code: 15:06:32 UTC,
java,
autosave
Code: 15:06:38 UTC,
cpp,
autosave
// you can use includes, for example:
// #include <algorithm>
#include <stack>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(string &S) {
// write your code in C++14 (g++ 6.2.0)
stack<char> s;
int count = 0;
if(S.empty())
return 1;
if(S.size() % 2 == 1 || S[0] != '(')
return 0;
for(int i = 0; i < S.size(); i++){
if(S[i] == '(')
s.push(S[i]);
else{
s.pop();
count++;
}
}
if(s.empty() && count == S.size() / 2)
return 1;
else
return 0;
}
Code: 15:08:31 UTC,
cpp,
autosave
// you can use includes, for example:
// #include <algorithm>
#include <stack>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(string &S) {
// write your code in C++14 (g++ 6.2.0)
stack<char> s;
int count = 0;
if(S.empty())
return 1;
if(S.size() % 2 == 1 || S[0] != '(')
return 0;
for(int i = 0; i < S.size(); i++){
if(S[i] == '(')
s.push(S[i]);
else{
s.pop();
count++;
}
}
if(s.empty() && count == S.size() / 2)
return 1;
else
return 0;
}
Code: 15:09:07 UTC,
cpp,
autosave
// you can use includes, for example:
// #include <algorithm>
#include <stack>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(string &S) {
// write your code in C++14 (g++ 6.2.0)
stack<char> s;
int count = 0;
if(S.empty())
return 1;
if(S.size() % 2 == 1 || S[0] != '(')
return 0;
for(int i = 0; i < S.size(); i++){
f
if(S[i] == '(')
s.push(S[i]);
else{
s.pop();
count++;
}
}
if(s.empty() && count == S.size() / 2)
return 1;
else
return 0;
}
Code: 15:09:29 UTC,
cpp,
verify,
result: Passed
// you can use includes, for example:
// #include <algorithm>
#include <stack>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(string &S) {
// write your code in C++14 (g++ 6.2.0)
stack<char> s;
int count = 0;
if(S.empty())
return 1;
if(S.size() % 2 == 1 || S[0] != '(')
return 0;
for(int i = 0; i < S.size(); i++){
if(s.empty() && S[i] == ')')
return 0;
if(S[i] == '(')
s.push(S[i]);
else{
s.pop();
count++;
}
}
if(s.empty() && count == S.size() / 2)
return 1;
else
return 0;
}
Analysis
Code: 15:09:32 UTC,
cpp,
final,
score: 
100
// you can use includes, for example:
// #include <algorithm>
#include <stack>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(string &S) {
// write your code in C++14 (g++ 6.2.0)
stack<char> s;
int count = 0;
if(S.empty())
return 1;
if(S.size() % 2 == 1 || S[0] != '(')
return 0;
for(int i = 0; i < S.size(); i++){
if(s.empty() && S[i] == ')')
return 0;
if(S[i] == '(')
s.push(S[i]);
else{
s.pop();
count++;
}
}
if(s.empty() && count == S.size() / 2)
return 1;
else
return 0;
}
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N)
expand all
Correctness tests
1.
0.001 s
OK
2.
0.001 s
OK
1.
0.001 s
OK
1.
0.001 s
OK
2.
0.001 s
OK
3.
0.001 s
OK
1.
0.001 s
OK
2.
0.001 s
OK
3.
0.001 s
OK
expand all
Performance tests
1.
0.001 s
OK
2.
0.001 s
OK
3.
0.001 s
OK
1.
0.004 s
OK
2.
0.001 s
OK
multiple_full_binary_trees
sequence of full trees of the form T=(TT), depths [1..10..1], with/without unmatched ')' at the end, length=49K+
sequence of full trees of the form T=(TT), depths [1..10..1], with/without unmatched ')' at the end, length=49K+
✔
OK
1.
0.001 s
OK
2.
0.001 s
OK
3.
0.001 s
OK
broad_tree_with_deep_paths
string of the form (TTT...T) of 300 T's, each T being '(((...)))' nested 200-fold, length=1 million
string of the form (TTT...T) of 300 T's, each T being '(((...)))' nested 200-fold, length=1 million
✔
OK
1.
0.012 s
OK
2.
0.001 s
OK